Vertices of a triangle are $\text{A}\left(3,2,1\right),\text{B}\left(1,-1,2\right)\text{and}\text{C}\left(2,1,1\right)$ . Area of the triangle will be :

$\overline{B}=\hat{i}-\hat{j}+2\hat{k}$

$\overline{C}=2\hat{i}+\hat{j}+\hat{k}$

$\overline{AB}=O\overline{A}-O\overline{B}=3\hat{i}+2\hat{j}+\hat{k}-\left(\hat{i}-\hat{j}+2\hat{k}\right)$

$=3\hat{i}+2\hat{j}+\hat{k}-\hat{i}+\hat{j}-2\hat{k}$

$\overline{AB}=2\hat{i}+3\hat{j}-\hat{k}$

$\overline{AC}=O\overline{A}-O\overline{C}=3\hat{i}+2\hat{j}+\hat{k}-\left(2\hat{i}+\hat{j}+\hat{k}\right)$

$=3\hat{i}+2\hat{j}+\hat{k}-2\hat{i}-\hat{j}-\hat{k}$

$\overline{AC}=\hat{i}+\hat{j}+0$

$\left|\overline{AB}\times \overline{AC}\right|=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& -1\\ 1& 1& 0\end{array}\right|$

$=\hat{i}\left(0+1\right)-\hat{j}\left(0+1\right)+\hat{k}\left(2-3\right)$

$=\hat{i}-\hat{j}+\hat{k}$

$=\sqrt{1+1+1}=\sqrt{3}$

$\therefore Area=\frac{1}{2}\left|\overline{AB}\times \overline{AC}\right|$

$=\frac{1}{2}\left(\sqrt{3}\right)$

$=\frac{\sqrt{3}}{2}$